The value of $$\cot^{-1}\left(\frac{\sqrt{1 + \tan^2(2)} - 1}{\tan(2)}\right) - \cot^{-1}\left(\frac{\sqrt{1 + \tan^2(\frac{1}{2})} + 1}{\tan(\frac{1}{2})}\right)$$ is equal to

Write the required value as

$$\cot^{-1}\!\left(\dfrac{\sqrt{1+\tan^{2}2}-1}{\tan 2}\right)\;-\;\cot^{-1}\!\left(\dfrac{\sqrt{1+\tan^{2}\!\left(\dfrac12\right)}+1}{\tan\dfrac12}\right)$$

The standard identity is $$\sqrt{1+\tan^{2}\theta}=|\sec \theta|$$. We must keep the modulus because $$\sec\theta$$ can be negative although the square root is always positive.

Case 1: $$\theta = 2 \text{ rad}$$ (first inverse-cot term)

Because $$\cos2\lt0$$, $$|\sec2|=-\sec2$$. Hence

$$\dfrac{\sqrt{1+\tan^{2}2}-1}{\tan2}= \dfrac{-\sec2-1}{\tan2}= -\dfrac{\sec2+1}{\tan2}$$

Now use the algebraic simplification

$$\dfrac{\sec x+1}{\tan x}= \dfrac{\dfrac1{\cos x}+1}{\dfrac{\sin x}{\cos x}}= \dfrac{1+\cos x}{\sin x}= \dfrac{2\cos^{2}\dfrac x2}{2\sin\dfrac x2\cos\dfrac x2}= \cot\dfrac x2$$

Putting $$x=2$$ gives $$\dfrac{\sec2+1}{\tan2}=\cot1$$, so

$$\dfrac{\sqrt{1+\tan^{2}2}-1}{\tan2}= -\cot1 = \cot\!\bigl(\pi-1\bigr)$$

(since $$\cot(\pi-\alpha)=-\cot\alpha$$). The principal branch of $$\cot^{-1}y$$ lies in $$(0,\pi)$$ and is monotonic, therefore

$$\cot^{-1}\!\left(\dfrac{\sqrt{1+\tan^{2}2}-1}{\tan2}\right)=\cot^{-1}\!\bigl(\cot(\pi-1)\bigr)=\pi-1$$

Case 2: $$\theta=\dfrac12 \text{ rad}$$ (second inverse-cot term)

Here $$\cos\dfrac12\gt0$$, so $$|\sec\dfrac12|=\sec\dfrac12$$. Thus

$$\dfrac{\sqrt{1+\tan^{2}\dfrac12}+1}{\tan\dfrac12}= \dfrac{\sec\dfrac12+1}{\tan\dfrac12}=\cot\dfrac14$$

Because $$0\lt \dfrac14\lt \dfrac\pi2$$, the principal value is simply

$$\cot^{-1}\!\left(\dfrac{\sqrt{1+\tan^{2}\dfrac12}+1}{\tan\dfrac12}\right)=\cot^{-1}\!\bigl(\cot\dfrac14\bigr)=\dfrac14$$

Final value

$$\bigl(\pi-1\bigr)-\dfrac14=\pi-\dfrac54$$

Therefore the required value equals $$\pi-\dfrac54$$, which corresponds to Option A.